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Last update: 31/07/2015

One of the most important points, building a cnc machine, is the project of a well dimensioned power supply. Normally we need of two different voltages: the first (usually 18-40V or more) to power the motors and another, usually 5V, to power the logic of the controllers.

In another article we have described how to obtain a clean 5V from a quite high voltage, we are going to describe now, very simply, how to realize a filtered power supply for any cnc machine controller.

Usually a cnc machine is characterized by the presence of, at least, three motors and normally these motor are driven by a controller using a PWM (pulse width modulation, also called chopper, usually from 1k to 20khz). This a method normally used to regulate the current flow inside the wires of the motor to obtain a costant current at low or high speeds, obtaining a lower temperature of the motor and general higher performancens.

Anyway we are here to talk about the power supply. In it's simplest form it is made by only three components:

• a transformer (regular or toroidal);
• a rectifier bridge;
• one or more filter capacitors (like the schematic on the left).

This is the minimum traditional, not switching, not stabilized, linear power supply used for any cnc machine.

The transformer reduce the alternate voltage from 100/220V to a lower voltage that must be rectified by the bridge and levelled by the capacitors.

Imagining the situation where we are using three 2A/phase stepper motors driven by a chopper controller. Avoiding specific calculation of the current in a chopper driver we can say that the real current flowing in the motor is 2/3 of the current in a non chopper driver. In the worst conditions we'll have current flowing in all the six phases of the three motors therefore the power supply must provide:

6 (phases) x 2 Amp * (2/3 chopper) = 8 Amperes.

Imaging we wish to power the CNC machine at 24V/8A. In this case we'll need a 18V trasformer because after the rectifier bridge we'll obtain

(18 * 1.41)-0.7 = 24.68V dued to the rectification factor.

The formula to calculate the minimum value of the filter capacitors is: 22000 / (Voltage/current) =  9.166uF (10000 uF); while to obtain a low ripple level use the following: C(Farad) = Amp/(2xHz*Voltage) where Hz is 50 or 60 according to the national standards, in our case C = 8/(120*24) = 27.777uF.  "In medio stat virtus" latins where used to say therefore we'll use a 20.000 uF capacitors (or two of 10.000)

To obtain a power supply of 24V-8A we need to purchase:

• a 180W (18v*10A) transformer (200W is the commercial size), slightly bigger to provide the necessary current;;
• a 30-40A bridge (yes, 30-40A just to prevents excessive heats - the price is almost identical);
• one capacitor (50-63V) of 20.000uF or two-three to reach the same level of capacity.

Another solution is the purchase of any commercial switching power supply :-).

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